On Coupled system of Nonlinear Ψ-Hilfer Hybrid Fractional Differential Equations

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Fractional Differential Equations

Ashwini D., Kishor Kucche, José Vanterler da C. Sousa

To cite this version:

Ashwini D., Kishor Kucche, José Vanterler da C. Sousa. On Coupled system of Nonlinear Ψ-Hilfer Hybrid Fractional Differential Equations. 2021. �hal-03189225�

Ashwini D. Mali¹ maliashwini144@gmail.com

Kishor D. Kucche² kdkucche@gmail.com J. Vanterler da C. Sousa³

vanterlermatematico@hotmail.com, jose.vanterler@edu.ufabc.br

1,2 Department of Mathematics, Shivaji University, Kolhapur-416 004, Maharashtra, India.

3 Centro de Matem´atica, Computa¸c˜ao e Cogni¸c˜ao, Universidade Federal do ABC, Avenida dos Estados, 5001, Bairro Bangu, 09.210-580, Santo Andr´e, SP - Brazil

Abstract

In this paper, we derive the existence of a solution for coupled system for hybrid fractional differential equation involving Ψ-Hilfer derivative for initial value problem and boundary value problem. In the view of an application, we provide an examples to exhibit the effectiveness of our achieved results.

Key words: Ψ-Hilfer fractional derivative; Fractional differential inequalities; Existence of solution; Extremal solutions; Comparison theorems.

2010 Mathematics Subject Classification: 34A38, 26A33, 34A12, 34A40.

1 Introduction

Motivated by the work of [35, 36], in the present paper, we consider the following Ψ-Hilfer hybrid FDEs of the form







HD^{µ, ν}₀₊ ^{; Ψ}

y(t)−w(t, y(t)) u(t, y(t))

=v

t, x(t), kI₀^µ₊^{; Ψ}x(t)

, a.e. t∈(0, T],

t→0+lim (Ψ (t)−Ψ (0))^1−ξy(t) =y0 ∈R^,

(1.1)

and







HD₀^{µ, ν}₊ ^{; Ψ}

x(t)−w(t, x(t)) u(t, x(t))

=v

t, y(t), kI₀^{µ; Ψ}₊ y(t)

, a.e. t∈(0, T],

t→0+lim (Ψ (t)−Ψ (0))^1−ξx(t) =y0∈R^,

(1.2)

where 0 < µ < 1, 0 ≤ ν ≤ 1, ξ = µ+ν(1−µ), ^HD^{µ,ν; Ψ}₀₊ (·) is the Ψ-Hilfer fractional derivative of order µand type ν, u∈C(J ×R^,R^{\ {0}), J = [0, T] and w∈C(J×}R^,R⁾ and v∈C(J×R^×R^,R^).

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications

1

[1, 3, 13, 17, 20, 24, 25]. The area of differential equations (integer and fractional order) dedicated to quadratic perturbations of nonlinear problems, also known as hybrid differential equations, is one of the most important areas and has attracted considerable attention from researchers. Over the past few decades, with the growth and greater demand for the theory of fractional differential equations, the search for discussing properties of solutions of hybrid differential problems, has gained prominence and greater investigation both in the theoretical sense and involving applications [5, 7, 9, 12, 14, 16]. Some fundamental and important papers involving the existence of solutions of fractional hybrid differential problems for further reading, we recommend the following works [14, 32, 35, 36].

In 2011, Zhao et al. [5], investigate the existence of solutions for fractional hybrid differential equations involving Riemann-Liouville differential operators of order 0< q <1





 D^q

x(t) f(t, x(t))

=g(t, x(t)), a.e. t∈J, x(0) = 0,

where f ∈ C(J ×R,R/{0}) and g ∈ C(J ×R,R), is proved under mixed Lipschitz and Caratheodory conditions.

Sun et al. [13], discuss the existence of solutions for the boundary value problem of fractional Hybrid differential equations





 D^α₀₊

x(t) f(t, x(t))

+g(t, x(t)) = 0, 0< t <1, x(0) =x(1) = 0,

where 1 < α ≤ 2 is a real number, D₀₊^α (·) is the Riemann-Liouville fractional derivative.

Other interesting works on fractional hybrid differential equations can be obtained in [4, 10, 11]. In 2017 Ali et al. [36], investigated sufficient conditions to discuss the existence of solutions for the study of coupled systems of fractional order hybrid.

We will here highlight the main objective discussed in this article, and its approach to the result, in order to facilitate and be clear, the main contribution.

We obtain, the equivalent fractional integral equation to the Ψ-Hilfer hybrid FDEs Eqs.(2.3)-(2.4) and establish the existence of solution in the weighted space C1−ξ; Ψ(J,R).

Our main objective here is to obtain estimates on Ψ-Hilfer derivative and utilize it to develop the hybrid fractional differential inequalities involving Ψ-Hilfer derivative. Using the fractional differential inequalities in the setting of Ψ-Hilfer derivative, we then derive the existence of extremal solutions and comparison results.

The obtained outcomes in the current paper additionally hold for nonlinear hybrid FDEs with well known fractional derivative operators recorded in [23], for example, RL, Caputo,Ψ- RL, Ψ-Caputo, Hadmard, Katugampola, Riesz, Erd´ely-Kober, Hilfer and so forth.

The plan of the paper is as follows: In the section 2, we give some definitions and results which are useful to prove the main results. In section 3, we derive an equivalent integral equation(IE) to the problem Eqs.(2.3)-(2.4) and establish existence of solution to it. In section 4, we obtain estimates on Ψ-Hilfer derivative. Section 5 deals with hybrid fractional differential inequalities involving Ψ-Hilfer derivative. Section 6, contribute the study of maximal and minimal solutions. Finally, comparison theorems are proved in the section 7.

2 Preliminaries

Let [a, b] (0< a < b <∞) be a finite interval and Ψ∈C¹([a, b],R) be an increasing function such that Ψ⁰(t)6= 0,∀ t∈[a, b]. We consider the weighted space

C1−ξ; Ψ[a, b] =n h

h: (a, b]→R, h(a+) exists and (Ψ (t)−Ψ (a))^1−ξh(t)∈C[a, b]o

, 0< ξ ≤1, endowed with the norm

khk_C

1−ξ;Ψ[a,b]= max

t∈[a,b]

(Ψ (t)−Ψ (a))^1−ξh(t)

. (2.1)

Definition 2.1 ([17]) Lethbe an integrable function defined on[a, b]. Then theΨ-Riemann- Liouville fractional integral of order µ >0 (µ∈R⁾ of the function h is given by

I_a^µ;Ψ+ h(t) = 1 Γ (µ)

Z t a

Ψ⁰(s) (Ψ (t)−Ψ (s))^µ−1h(s)ds. (2.2) Definition 2.2 ([23]) TheΨ-Hilfer fractional derivative of a functionhof order0< µ <1 and type 0≤ν ≤1, is defined by

HD^{µ, ν; Ψ}_a+ h(t) =I_a^{ν(1−µ); Ψ}+

1 Ψ⁰(t)

d dt

I(1−ν)(1−µ); Ψ

a⁺ h(t).

Lemma 2.1 ([17, 23]) Let χ, δ >0 and ρ > n. Then (i) I_a^µ+^{; Ψ}I_a^χ+^{; Ψ}h(t) =I_a^µ+χ+ ^{; Ψ}h(t).

(ii) I_a^µ₊^{; Ψ}(Ψ (t)−Ψ (a))^δ−1= Γ (δ)

Γ (µ+δ)(Ψ (t)−Ψ (a))^µ+δ−1. (iii) ^HD^{µ, ν}_a+ ^{; Ψ}(Ψ (t)−Ψ (a))^ξ−1 = 0.

(iv) ^HD^{µ, ν}_a+ ^{; Ψ}(Ψ (t)−Ψ (a))^ρ−1= Γ (ρ)

Γ (ρ−µ)(Ψ (t)−Ψ (a))^µ−ρ−1. Lemma 2.2 ([23]) If h∈Cⁿ[a, b], n−1< µ < nand 0≤ν ≤1, then

(i) I_a^µ;Ψ+ HD_a^µ,ν+ ^;Ψh(t) =h(t)−

n

P

k=1

(Ψ (t)−Ψ (a))^ξ−k

Γ (ξ−k+ 1) h^[n−k]_Ψ I(1−ν)(n−µ) ;Ψ

a⁺ h(a)where,h^[n−k]_Ψ h(t) = 1

Ψ⁰(t) d dt

n−k

h(t).

(ii) ^HD^µ,ν_a+ ^;ΨI_a^µ+^;Ψh(t) =h(t).

Lemma 2.3 ([34]) Let 0< µ <1, 0≤ν ≤1, ξ =µ+ν(1−µ), f ∈C(J×R^,R^{\ {0}) is} bounded,J = [0, T]andg∈C(J×R^,R^{) ={h|the mapω→h(τ, ω)}is continuous for eachτ and the map τ → h(τ, ω) is measurable for each ω}. A function y ∈ C_{1−ξ; Ψ}(J, R^{) is the} solution of hybrid FDEs

HD^{µ, ν}₀₊ ^{; Ψ}

y(t) f(t, y(t))

=g(t, y(t)), a.e. t∈(0, T], (2.3)

(Ψ (t)−Ψ (0))^1−ξy(t)|_t=0 =y0 ∈R^{, (2.4)} if and only if it is solution of the following hybrid fractional integral equation (IE)

y(t) =f(t, y(t))

y₀

f(0, y(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ₊^{; Ψ}g(t, y(t))

, t∈(0, T]. (2.5)

Definition 2.3 ([37]) An element (x, y) ∈ X ×X is called a coupled fixed point of a mapping T :X×X→X if T(x, y) =x and T(y, x) =y.

Lemma 2.4 ([35]) LetS be a non-empty, closed, convex and bounded subset of the Banach algebra X and S˜=S×S. Suppose that E, G:X →X andF :S →X are three operators such that

(a) E and G are Lipschitzian with a Lipschitz constantsσ and δ respectively;

(b) F is completely continuous;

(c) y=Ey F x+Gy =⇒ y∈S for allx∈S and (d) 4σM+δ <1 where M = sup{kBxk:x∈S}.

Then, the operator equationT(y, x) =Ey F x+Gy has a at least one coupled fixed point in S.˜

Lemma 2.5 ([38]) Let S^∗ be a non-empty, closed, convex and bounded subset of the Ba- nach space E and let A, C :E→E and B :S^∗ →E are three operators such that

(a) A and C are Lipschitzian with a Lipschitz constantsK and L respectively;

(b) B is completely continuous;

(c) y=Ay Bx+Cy =⇒ y∈S^∗ for all x∈S^∗ and (d) K M^∗+ L <1 where M^∗ = sup{kByk:y∈S^∗}.

Then, the operator equation Ay By+Cy=y has a solution inS^∗.

3 IVP for Coupled system of Hyrid FDEs

An application of the Lemma 2.3 gives the equivalent fractional IE to the FDEs Eq.(1.1), given in the following Lemma.

Lemma 3.1 A function y∈C1−ξ; Ψ(J,R)is the solution of the Cauchy problem for hybrid FDEs Eq.(1.1)if and only if it is solution of the following hybrid fractional IE

y(t) =u(t, y(t))

y0

u(0, y(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ₊^{; Ψ}v

t, x(t), kI₀^{µ; Ψ}₊ x(t)

+w(t, y(t)), t∈(0, T]. (3.1)

We list the following assumptions to prove the existence of solution to the coupled system of hybrid FDEs Eqs.(1.1)-(1.2).

(H1) The functions u ∈ C(J×R^,R^{\ {0}) and w ∈ C(J×}R^,R) are bounded and there exists constants σ, δ >0 such that for all p, q∈R^{and t∈J = [0, T] we have}

|u(t, p)−u(t, q)| ≤σ|p−q|

and

|w(t, p)−w(t, q)| ≤δ|p−q|.

(H2) The function v ∈C(J×R^×R^,R) and there exists a functiong ∈ C1−ξ(J,R) such that

|v(t, p, q)| ≤(Ψ (t)−Ψ (0))^1−ξg(t), a.e. t∈J and p, q∈R^.

Theorem 3.2 Assume that the hypotheses (H1)-(H2) hold. Then, the coupled system of nonlinear Ψ-Hilfer hybrid FDEs Eqs.(1.1)-(1.2) has a solution (y, x) ∈ C1−ξ; Ψ(J,R)× C1−ξ; Ψ(J,R) provided

4σ (

y0

u(0, y(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ(µ+ 1) kgk_C

1−ξ; Ψ(J,R)

)

+δ <1. (3.2)

Proof: Let X :=

C1−ξ; Ψ(J,R),k·k_C

1−ξ; Ψ(J,R)

. Then X is a Banach algebra with the product of vectors defined by (xy)(t) =x(t)y(t), t∈(0, T]. Define,

S ={x∈X:kxk_C

1−ξ; Ψ(J,R) ≤R}, where

R=K1

(

y₀ u(0, x(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ(µ+ 1) kgk_C

1−ξ; Ψ(J,R)

)

+K2 (Ψ (T)−Ψ (0))^1−ξ and K₁ >0 andK₂ >0 are the constants such that |u(t,·)|< K₁ and |w(t,·)|< K₂ for all t.

Clearly,S is non-empty, closed, convex and bounded subset ofX. If (y, x)∈S×S = ˜S is a solution of the coupled system of nonlinear Ψ-Hilfer hybrid FDEs Eqs.(1.1)-(1.2), then

(y, x)∈S×S = ˜S is a solution of the coupled system of fractional IEs



















y(t) =u(t, y(t))

y0

u(0, y(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ₊^{; Ψ}v

t, x(t), kI₀^µ₊^{; Ψ}x(t)

+w(t, y(t)), x(t) =u(t, x(t))

y0

u(0, x(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ+^{; Ψ}v

t, y(t), kI₀^µ+^{; Ψ}y(t)

+w(t, x(t)), t∈(0, T].

(3.3)

Define three operatorsE, G:X→X andF :S→X by Ey(t) =u(t, y(t)), t∈J;

F y(t) = y0

u(0, y(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ₊^{; Ψ}v

t, y(t), kI₀^{µ; Ψ}₊ y(t)

, t∈(0, T];

Gy(t) =w(t, y(t)), t∈J.

Then, the coupled hybrid IEs Eq.(3.3) transformed into the coupled system of operator equations as

(y =Ey F x+Gy, y∈X

x =Ex F y+Gx, x∈X. (3.4)

Consider the mapping T : ˜S→X, S˜=S×S defined by T(y, x) =Ey F x+Gy, (y, x)∈S.˜

Then the coupled system of operator equations (3.4) can be written as y=T(y, x) and x=T(x, y), (y, x),(x, y)∈S.˜

To prove that the mappingT has coupled fixed point, we show that the operatorsE,F and G satisfies all the conditions of Lemma 2.4. The proof is given in the several steps:

Step 1: E, G:X→X are Lipschitz operators.

Using the hypothesis (H1), we obtain

(Ψ (t)−Ψ (0))^1−ξ(Ex(t)−Ey(t)) =

(Ψ (t)−Ψ (0))^1−ξ(u(t, x(t))−u(t, y(t)))

≤σ

(Ψ (t)−Ψ (0))^1−ξ(x(t)−y(t))

≤σkx−yk_C

1−ξ; Ψ(J,R). This gives,

kEx−Eyk_C

1−ξ; Ψ(J,R)≤σkx−yk_C

1−ξ; Ψ(J,R).

Therefore, E is Lipschitz operator with Lipschitz constant σ. On the similar line one can verify that that G is Lipschitz operator. Let δ is Lipschitz constant corresponding to operatorG.

Step 2: F :S→X is completely continuous.

(i) F :S →X is continuous.

Let {y_n} be any sequence in S such that y_n → y as n → ∞ in S. We prove that F y_n→F y asn→ ∞ inS. Consider,

kF y_n−F yk_C

1−ξ; Ψ(J,R) = max

t∈J

(Ψ (t)−Ψ (0))^1−ξ(F y_n(t)−F y(t))

≤max

t∈J

(Ψ (t)−Ψ (0))^1−ξ Γ (µ)

Z t 0

Ψ⁰(s)(Ψ(t)−Ψ(s))^µ−1×

v

s, y_n(s), kI₀^{µ; Ψ}+ y_n(s)

−v

s, y(s), kI₀^µ+^{; Ψ}y(s) ds.

By continuity of v and Lebesgue dominated convergence theorem, from the above in- equality, we obtain

kF y_n−F yk_C

1−ξ; Ψ(J,R)→0 asn→ ∞.

This provesF :S→X is continuous.

(ii) F(S) ={F y:y∈S}is uniformly bounded.

Using hypothesis (H2), for anyy∈S andt∈J, we have

(Ψ (t)−Ψ (0))^1−ξF y(t)

≤

y₀ u(0, y(0+))

+ (Ψ (t)−Ψ (0))^1−ξ Γ (µ)

Z t 0

Ψ⁰(s)(Ψ(t)−Ψ(s))^µ−1 v

s, y(s), kI^{µ; Ψ}

0⁺ y(s) ds

≤

y0

u(0, y(0+))

+ (Ψ (t)−Ψ (0))^1−ξ Γ (µ)

Z t 0

Ψ⁰(s)(Ψ(t)−Ψ(s))^µ−1(Ψ(s)−Ψ(0))^1−ξg(s)ds

≤

y₀ u(0, y(0+))

+kgk_C

1−ξ; Ψ(J,R) (Ψ (t)−Ψ (0))^1−ξ(Ψ (t)−Ψ (0))^µ Γ (µ+ 1)

≤

y₀ u(0, y(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ (µ+ 1) kgk_C

1−ξ; Ψ(J,R). Therefore,

kF yk_C

1−ξ; Ψ(J,R) ≤

y₀ u(0, y(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ (µ+ 1) kgk_C

1−ξ; Ψ(J,R). (3.5) (iii) F(S) is equicontinuous.

Let anyy ∈S andt₁, t₂∈J witht₁ < t₂. Then using hypothesis(H2), we have

(Ψ (t₂)−Ψ (0))^1−ξF y(t₂)−(Ψ (t₁)−Ψ (0))^1−ξF y(t₁)

=

( y₀

u(0, y(0+)) +(Ψ (t₂)−Ψ (0))^1−ξ Γ (µ)

Z t2

0

Ψ⁰(s)(Ψ(t₂)−Ψ(s))^µ−1v

s, y(s), kI₀^µ₊^{; Ψ}y(s) ds

)

−

( y₀

u(0, y(0+))+ (Ψ (t₁)−Ψ (0))^1−ξ Γ (µ)

Z t1

0

Ψ⁰(s)(Ψ(t₁)−Ψ(s))^µ−1v

s, y(s), kI^{µ; Ψ}

0⁺ y(s) ds

)

≤

(Ψ (t2)−Ψ (0))^1−ξ Γ (µ)

Z t2

0

Ψ⁰(s)(Ψ(t2)−Ψ(s))^µ−1 v

s, y(s), kI₀^µ₊^{; Ψ}y(s)

ds

−(Ψ (t1)−Ψ (0))^1−ξ Γ (µ)

Z t1

0

Ψ⁰(s)(Ψ(t1)−Ψ(s))^µ−1 v

s, y(s), kI₀^µ₊^{; Ψ}y(s)

ds

≤

(Ψ (t2)−Ψ (0))^1−ξ Γ (µ)

Z t2

0

Ψ⁰(s)(Ψ(t2)−Ψ(s))^µ−1(Ψ(s)−Ψ(0))^1−ξg(s)ds

−(Ψ (t1)−Ψ (0))^1−ξ Γ (µ)

Z _t1

0

Ψ⁰(s)(Ψ(t₁)−Ψ(s))^µ−1(Ψ(s)−Ψ(0))^1−ξg(s)ds

≤

(Ψ (t₂)−Ψ (0))^1−ξkgk_C

1−ξ; Ψ(J,R)

Γ (µ)

Z t2

0

Ψ⁰(s)(Ψ(t₂)−Ψ(s))^µ−1ds

−(Ψ (t1)−Ψ (0))^1−ξkgk_C

1−ξ; Ψ(J,R)

Γ (µ)

Z t1

0

Ψ⁰(s)(Ψ(t1)−Ψ(s))^µ−1ds

= kgk_C

1−ξ; Ψ(J,R)

Γ (µ+ 1) n

(Ψ(t₂)−Ψ(0))^µ+1−ξ−(Ψ(t₁)−Ψ(0))^µ+1−ξo . By the continuity of Ψ, from the above inequality it follows that

if|t₁−t2| →0 then

(Ψ (t2)−Ψ (0))^1−ξF y(t2)−(Ψ (t1)−Ψ (0))^1−ξF y(t1) →0.

From the parts (ii) and (iii), it follows thatF(S) is uniformly bounded and equicontinous set inX. Then by Arzel´a-Ascoli theorem,F(S) is relatively compact. We have proved that, F :S→Xis a compact operator. SinceF :S →Xis the continuous and compact operator, it is completely continuous.

Step 3: For y∈X, y=Ey F x+Gy =⇒ y∈S for all x∈S .

Let any y ∈X and x∈S such that y=Ey F x+Gy. Using the hypothesis (H2) and boundedness of u and w, for anyt∈J, we have

(Ψ (t)−Ψ (0))^1−ξy(t)

=

(Ψ (t)−Ψ (0))^1−ξ[Ey(t)F x(t) +Gy(t)]

=

(Ψ (t)−Ψ (0))^1−ξ

u(t, y(t))

y0

u(0, x(0+))(Ψ (t)−Ψ (0))^ξ−1+I₀^µ₊^{; Ψ}v

t, x(t), kI₀^µ₊^{; Ψ}x(t)

+w(t, y(t))

=

u(t, y(t))

y₀

u(0, x(0+)) + (Ψ (t)−Ψ (0))^1−ξI₀^µ₊^{; Ψ}v

t, x(t), kI₀^{µ; Ψ}₊ x(t)

+ (Ψ (t)−Ψ (0))^1−ξw(t, y(t))

≤ |u(t, y(t))|

(

y₀ u(0, x(0+))

+(Ψ (t)−Ψ (0))^1−ξ Γ (µ)

Z t 0

Ψ⁰(s)(Ψ(t)−Ψ(s))^µ−1 v

s, x(s), kI₀^µ₊^{; Ψ}x(s) ds

)

+ (Ψ (t)−Ψ (0))^1−ξ|w(t, y(t))|

≤K₁ (

y₀ u(0, x(0+))

+(Ψ (t)−Ψ (0))^1−ξ Γ (µ)

Z t 0

Ψ⁰(s)(Ψ(t)−Ψ(s))^µ−1(Ψ (s)−Ψ (0))^1−ξg(s)ds )

+ (Ψ (t)−Ψ (0))^1−ξK2

≤K1

(

y0

u(0, x(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ (µ+ 1) kgk_C

1−ξ; Ψ(J,R)

)

+ (Ψ (T)−Ψ (0))^1−ξK2. This gives

kyk_C

1−ξ; Ψ(J,R)≤K1

(

y0

u(0, x(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ kgk_C

1−ξ; Ψ(J,R)

Γ (µ+ 1)

)

+ (Ψ (T)−Ψ (0))^1−ξK₂=R.

This implies, y∈S.

Step 4: To prove 4σM +δ <1 where M = supn kF yk_C

1−ξ; Ψ(J,R):y∈So . From inequality (3.5), we have

M = sup n

kF yk_C

1−ξ; Ψ(J,R) :y∈S o

≤

y0

u(0, y(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ (µ+ 1) kgk_C

1−ξ; Ψ(J,R). Now, using the condition (3.2), we have

4σM +δ≤4σ (

y0

u(0, y(0+))

+(Ψ (T)−Ψ (0))^µ+1−ξ Γ (µ+ 1) kgk_C

1−ξ; Ψ(J,R)

)

+δ <1.

From Steps 1 to 4, it follows that all the conditions of Lemma 2.4 are fulfilled. Con- sequently, by applying Lemma 2.4, the operator T has a coupled solution in ˜S = S ×S.

Hence, the coupled system of hybrid FDEs Eqs.(1.1)-(1.2) has a solution in C1−ξ; Ψ(J,R).

2

4 BVPs for Coupled system of Hybrid FDEs

In this section, we are concerned with the following BVPs for coupled system of hybrid FDEs of the form







HD₀^{µ, ν}₊ ^{; Ψ}

y(t)−w₁(t, y(t), x(t)) u₁(t, y(t), x(t))

=v₁(t, y(t), x(t)), a.e. t∈(0, T], a lim

t→0+(Ψ (t)−Ψ (0))^1−ξy(t) +b lim

t→T(Ψ (t)−Ψ (0))^1−ξy(t) =y0∈R^,

(4.1) and







HD^{µ, ν}₀₊ ^{; Ψ}

x(t)−w2(t, y(t), x(t)) u₂(t, y(t), x(t))

=v2(t, y(t), x(t)), a.e. t∈(0, T], a lim

t→0+(Ψ (t)−Ψ (0))^1−ξx(t) +b lim

t→T(Ψ (t)−Ψ (0))^1−ξx(t) =y0∈R^,

(4.2)

where 0 < µ < 1, 0 ≤ ν ≤ 1, ξ = µ+ ν(1−µ)(0 < ξ ≤ 1),^HD^µ,ν₀₊^{; Ψ}(·) is the Ψ- Hilfer fractional derivative of order µ and type ν, J = [0, T], a 6= 0 and b 6= 0 are the constants, ui ∈ C(J ×R^×R^,R^{\ {0})(i = 1,2), wi ∈ C(J ×}R^×R^,R^{)(i = 1,2) and} vi ∈C(J ×R^×R^,R^{)(i= 1,2).}

Consider the product space E=X×X, X =C1−ξ;Ψ(J,R) with

(i) vector addition: (p, q)(t) + (¯p,q)(t) = (p(t) + ¯¯ p(t), q(t) + ¯q(t)) ;

(ii) scalar multiplication: k(p, q)(t) = (k p(t), k q(t)), where,p, q∈X and k∈R^. Then, E is a Banach algebra with the norm

k(p, q)k_E =kpk_C

1−ξ;Ψ(J,R)+kqk_C

1−ξ;Ψ(J,R) (4.3)

and the vector multiplication defined by

(p, q)(t)·(¯p,q)(t) = (p(t)¯¯ p(t), q(t)¯p(t)),for any (p, q),(¯p,q)¯ ∈E and t∈J.

Theorem 4.1 The BVP for Ψ-Hilfer hybridFDEs

HD^{µ, ν}₀₊ ^{; Ψ}

y(t)−w1(t, y(t), x(t)) u1(t, y(t), x(t))

=v1(t, y(t), x(t)), a.e. t∈(0, T], (4.4) a lim

t→0+(Ψ (t)−Ψ (0))^1−ξy(t) +b lim

t→T(Ψ (t)−Ψ (0))^1−ξy(t) =y0 ∈R^{, (4.5)} is equivalent to the fractional IE

y(t) =w₁(t, y(t), x(t)) +u₁(t, y(t), x(t))h

(Ψ (t)−Ψ (0))^ξ−1Ω₁+I^{µ; Ψ}

0⁺ v₁(t, y(t), x(t))i

, t∈(0, T], (4.6) where

Ω1=

y₀−b (Ψ (T)−Ψ (0))^1−ξ

w₁(T, y(T), x(T)) +u₁(T, y(T), x(T))I₀^{µ; Ψ}+ v₁(T, y(T), x(T)) a u₁(0, y(0+), x(0+)) +b u₁(T, y(T), x(T)) . Proof: Lety∈C1−ξ(J,R) is a solution of the BVP for Ψ-Hilfer hybrid FDEs Eqs.(4.4)- (4.5). TakingI₀^{µ; Ψ}₊ on both sides of Eq.(4.4) and using Lemma 2.2 (i), we get

y(t)−w₁(t, y(t), x(t))

u1(t, y(t), x(t)) −(Ψ (t)−Ψ (0))^ξ−1 Γ(ξ)

I₀^{1−ξ; Ψ}+

y(t)−w₁(t, y(t), x(t)) u1(t, y(t), x(t))

t=0

=I₀^{µ; Ψ}+ v1(t, y(t), x(t))

Let C^∗ =

I₀^{1−ξ; Ψ}+

y(t)−w₁(t, y(t), x(t)) u1(t, y(t), x(t))

t=0

. Thus, we have y(t)−w₁(t, y(t), x(t))

u1(t, y(t), x(t)) = (Ψ (t)−Ψ (0))^ξ−1

Γ(ξ) C^∗+I₀^{µ; Ψ}+ v1(t, y(t), x(t)).

Therefore,

y(t) =w₁(t, y(t), x(t)) +u₁(t, y(t), x(t))

"

(Ψ (t)−Ψ (0))^ξ−1

Γ(ξ) C^∗+I^{µ; Ψ}

0⁺ v₁(t, y(t), x(t))

# . (4.7)